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Question

Get the factorisation (a+b+c)3a3b3c3=3(a+b)(b+c)(c+a) writing the expression
(a+b+c)3a3b3c3=[(a+b+c)3a3][b3+c3]

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Solution

(a+b+c)3a3b3c3
=[(a+b+c)3a3][b3+c3]
={(a+b+ca)[(a+b+c)2+a2+(a+b+c)a]}[(b+c)(b2+c2bc)]
={(b+c)[(a+b+c)2+a2+(a+b+c)a]}[(b+c)(b2+c2bc)]
=(b+c)[(a+b+c)2+a2+a2+ab+acb2c2+bc] =(b+c)[(a2+b2+c2+2ab+2bc+2ac)+2a2+ab+acb2c2+bc]
=(b+c)[3a2+3ab+3bc+3ac]
=3(b+c)[a2+ab+bc+ac]
=3(b+c)[a(a+b)+c(b+a)]
=3(b+c)(a+b)(a+c)
Hence, (a+b+c)3a3b3c3=3(a+b)(b+c)(c+a)

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